A cube with side length $P ext{ cm}$ is inscribed inside a sphere of radius $R ext{ cm}$ such that the sphere touches all the vertices of the cube. Find the ratio of the total surface area of the sphere to the lateral surface area of the cube.

Question

A cube with side length $P 	ext{ cm}$ is inscribed inside a sphere of radius $R 	ext{ cm}$ such that the sphere touches all the vertices of the cube. Find the ratio of the total surface area of the sphere to the lateral surface area of the cube.

Accepted Answer

**Solution:**

1. Since the cube is placed inside the sphere such that the sphere touches all vertices of the cube, the diagonal of the cube equals the diameter of the sphere.

2. The diagonal of the cube is given by:
   $$\text{Diagonal} = P \sqrt{3}$$

3. The diameter of the sphere is:
   $$2R$$

4. Equate the diagonal of the cube to the diameter of the sphere:
   $$P \sqrt{3} = 2R \implies P = \frac{2R}{\sqrt{3}}$$

5. Total surface area of the sphere:
   $$4 \pi R^2$$

6. Lateral surface area of the cube is the sum of areas of the four vertical faces (excluding top and bottom):
   $$4 \times P^2$$

7. Required ratio:
   $$
   \frac{\text{Surface area of sphere}}{\text{Lateral surface area of cube}} = \frac{4 \pi R^2}{4 P^2} = \frac{\pi R^2}{P^2}
   $$

8. Substitute $P = \frac{2R}{\sqrt{3}}$:
   $$
   \frac{\pi R^2}{\left(\frac{2R}{\sqrt{3}}\right)^2} = \frac{\pi R^2}{\frac{4R^2}{3}} = \frac{\pi R^2 \times 3}{4 R^2} = \frac{3\pi}{4}
   $$

9. Using $\pi = \frac{22}{7}$:
   $$
   \frac{3 \times \frac{22}{7}}{4} = \frac{66}{28} = \frac{33}{14}
   $$

**Answer:** $33 : 14$

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